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Dec 17, 1998, 3:00:00 AM12/17/98

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Hi all,

This is kind of an embarrassing question to ask, since I *know* I learned

this in high school. But then, that was a long time ago. So here goes:

On a 2D plane, given a point (x,y) and a slope (m), how can I extend a line

with a length of (n) units along the slope, and find the endpoints (x1,y1)

of that line?

Many thanks in advance.

-Matt

Dec 17, 1998, 3:00:00 AM12/17/98

to

Matt:

I am assuming that you want to find the OTHER endpoint which is N units

away from X1,Y1 at a slope M.

Two rules are helpful:

(x-x1)^2 + (y-y1)^2 = N^2 => N squared equals the X distance squared

plus the Y distance squared.

(y-y1)/(x-x1)=M => The change in Y over the change in X equals the

slope.

or

(y-y1)=(x-x1)*M => This can then be substituted into the first

equation to solve for X:

(x-x1)^2 * (1+m^2) = N^2

Solve for X, plug into the slope equation, and get Y.

Hope this helps.

Dec 17, 1998, 3:00:00 AM12/17/98

to

Hi,

The simplest way, in my opinion, is to take the slope, we'll use 4/3 for

example, and start at the given point, and go up 4, since that is the top

number (if it were -4/3 you would go down instead of up) and right 3, since

that is the bottom number (It won't be negetive.) Hope this helps.

The simplest way, in my opinion, is to take the slope, we'll use 4/3 for

example, and start at the given point, and go up 4, since that is the top

number (if it were -4/3 you would go down instead of up) and right 3, since

that is the bottom number (It won't be negetive.) Hope this helps.

EggyDee

Matt Kimmel wrote in message <75b9vk$3se$1...@winter.news.rcn.net>...

Dec 17, 1998, 3:00:00 AM12/17/98

to

"Matt Kimmel" <mki...@cyberlore-studios.com> wrote:

> On a 2D plane, given a point (x,y) and a slope (m), how can I extend a

> line with a length of (n) units along the slope, and find the endpoints

> (x1,y1) of that line?

If you find a unit vector ( a, b ) in the right direction, the solution

will be:

( x, y ) + n ( a, b )

in other words:

( x + n*a , y + n*b )

In fact a,b are not difficult to find if you have _any_ vector in the

right direction. For example

( 1, m ) with length sqrt(1 + m^2)

==> a = 1/(sqrt(1 + m^2)) and b = m/(sqrt(1 + m^2))

Ken, __O

_-\<,_

(_)/ (_)

Virtuale Saluton.

Dec 18, 1998, 3:00:00 AM12/18/98

to

In newsgroup alt.algebra.help, article

<75b9vk$3se$1...@winter.news.rcn.net>, the lovely and talented

<75b9vk$3se$1...@winter.news.rcn.net>, the lovely and talented

mki...@cyberlore-studios.com wrote:

> On a 2D plane, given a point (x,y) and a slope (m), how can I extend a line

> with a length of (n) units along the slope, and find the endpoints (x1,y1)

> of that line?

> On a 2D plane, given a point (x,y) and a slope (m), how can I extend a line

> with a length of (n) units along the slope, and find the endpoints (x1,y1)

> of that line?

This is a vector problem. The solution is a vector of length n, at slope

m. The slope is the tangent of the angle from the x-axis, so the vector's

direction angle is arctan(m). But the problem can be solved without

trigonometry.

Draw a little picture, which will make the next bit easier. You have a

right triangle. The x-axis is parallel to one leg; the point (x,y) is one

vertex, the desired line (the vector) is the hypotenuse, and therefore

(x1,y1) is the other end of the hypotenuse. Drop a perpendicular to where

it meets the other leg.

Now the hypotenuse measures n. The legs measure (x1-x) and (y1-y), and

you know (given the slope) that (y1-y) = m(x1-x) or y1 = m(x1-x)+y. Now

how to solve for x1? Well, you also know that (x1-x)^2+(y1-y)^2=n^2

(theorem of Pythagoras), so you can solve that way. It's a bit messy, but

that's an artifact of the way the problem was stated.

In trig terms, it's a lot easier. Using the same picture, define angle A

such that tan(A) = m. Then x1 = x+n*cos(A), y1 = y+n*sin(A).

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